4u Mathematics Marathon V 1.0 (1 Viewer)

[pLuvia]

Q: If I_n = ∫₀¹(1-x²)ⁿ dx, find I_n in terms of I_n-1 and n.

Spoiler

int.{0 to 1}(1-x²)ⁿ dx
Let u=(1-x²)ⁿdx dv=dx
du=-2xn(1-x²)^n-1dx v=x
I_n
=int.{0 to 1}(1-x²)ⁿ dx
=[x(1-x²)ⁿ]{0 to 1}+int.{0 to 1}n2x²(1-x²)^n-1dx
=2n int.{0 to 1}x²(1-x²)^n-1dx
=2n int.{0 to 1}-(1-x²-1)(1-x²)^n-1dx
=-2n int.{0 to 1}(1-x²)ⁿ-(1-x²)^n-1dx
=-2n[I_n-I_n-1]
I_n(1+2n)=2nI_n-1
I_n=2nI_n-1/(2n+1)

 

[pLuvia]

Opps, I'll edit it
Yeh forgot to put it in after the first IBP

 

[pLuvia]

Just put an easy one to get this thread bumped up again

Prove that
[a+b+c]/3>√abc

 

[Riviet]

Any restrictions on a, b and c?

 

[pLuvia]

a,b,c are real positive numbers

 

[pLuvia]

Prove a/b + b/c + c/a + c/b + b/a + a/c > 6

Spoiler

a/b+b/c+c/a+c/b+b/a+a/c-6
=[a²c+ab²+bc²+ac²+b²c+a²b-6abc]/abc
=[1/abc][a(b²-2bc+c²)+b(a²-2ac+c²)+c(a²-2ab+b²)]
=[1/abc][a(b-c)²+b(a-c)²+c(a-b)²]
>0
Hence a/b+b/c+c/a+c/b+b/a+a/c>6

 

[Mountain.Dew]

Spoiler

We know that a² + b² > 2ab.

Therefore a + b > 2√ab by substitution.

and similarly b + c > 2√bc and a + c > 2√ac

So if we add all these together we get 2 (a + b + c) > 6√abc

Therefore a + b + c > 3√abc.

whats the 'proof' for the fact that 2√ac + 2√bc + 2√ac > 6√abc?

is it assumed that √ac or √bc or √ab > 1?

Prove a/b + b/c + c/a + c/b + b/a + a/c > 6

Spoiler

a/b+b/c+c/a+c/b+b/a+a/c-6
=[a²c+ab²+bc²+ac²+b²c+a²b-6abc]/abc
=[1/abc][a(b²-2bc+c²)+b(a²-2ac+c²)+c(a²-2ab+b²)]
=[1/abc][a(b-c)²+b(a-c)²+c(a-b)²]
>0
Hence a/b+b/c+c/a+c/b+b/a+a/c>6

ooooo very elegant solution

 

[pLuvia]

ooooo very elegant solution

What's the other way of doing it? Just wanting to know

 

[pLuvia]

Extension to the question
First prove that [a+b+c+d]/4=⁴√abcd
Then find an expression for d in terms of a,b and c

That should help

 

[Yip]

(a+b)/2>√ab
(c+d)/2>√cd
(a+b+c+d)/2>√ab+√cd
a+b+c+d>2(√ab+√cd)>2(2√√abcd)=4⁴√abcd
[a+b+c+d]/4=⁴√abcd

Letting d=a+b+c/3,

LHS=4/3(a+b+c)>4⁴√abc(⁴√[a+b+c]/⁴√3)
[(a+b+c)^(3/4)]/3>⁴√abc/⁴√3
[(a+b+c)^(3/4)]>⁴√27abc
(a+b+c)^3>27abc
a+b+c>3³√abc

Prove that
[a+b+c]/3>√abc

I think pluvia had a typo,
if u sub a=1,b=2,c=3, u get 2>√6 which is obviously false
2>³√6 makes more sense

UMAT tomorrow -_- nande...~.~

 

[Mountain.Dew]

What's the other way of doing it? Just wanting to know

we know that:

a² + b² > 2ab

therefore, when divide a/b + b/a > 2

hence, b/c + c/b > 2, c/a + a/c > 2

add em all together: a/b + b/c + c/a + c/b + b/a + a/c > 6

ummm i take back my comment on your 'elegant solution'. i find this far easier and more efficient sorry, tis was my bad, pLuvia.

 

[Riviet]

Ooh nice proof Yip, also good luck with UMAT too, don't stay up too late studying for it!

 

[pLuvia]

Opps my bad, I actually meant to type
[a+b+c/3]³>√abc
Sorry LottoX

Good luck with the UMAT

 

[bboyelement]

S'ok, as long as I know it's not my crapiness making me fail the question.

I can't believe the UMAT is actually tomorrow. I feel so nervous.

Anyway:

Given: t/(a+b) = 1, prove 1/a² + 1/b² > 8/t²

therefore t=a+b
1/a²+ 1/b² = (1/a - 1/b)² + 2/ab
1/a²+ 1/b² > 2/ab
(a + b)² = (a - b)²+ 4ab
(a + b)² > 4ab
ab< (a + b)²/4

1/a² + 1/b² > 2/(a + b)²/4
1/a² + 1/b² > 8/(a + b)²
1/a² + 1/b² > 8/t²

prove
(b + c + d)(a + c + d)(a + b + d)(a + b + c) > 81abcd

btw. a,b,c,d all bigger than 0

 

[Riviet]

prove
(b + c + d)(a + c + d)(a + b + d)(a + b + c) > 81abcd

btw. a,b,c,d all bigger than 0

By AM/GM inequality:
(b+c+d)/3 > (bcd)^1/3 (1)

Similarly,
(a+c+d)/3 > (acd)^1/3 (2)

(a+b+d)/3 > (abd)^1/3 (3)

(a+b+c)/3 > (abc)^1/3 (4)

(1)(2)(3)(4) => {(b+c+d)(a+c+d)(a+b+d)(a+b+c)}/81 > (a³b³c³d³)^1/3

.'. (b+c+d)(a+c+d)(a+b+d)(a+b+c) > 81abcd, as required.

 

[Riviet]

Next Question:
The region bounded by y=1/(x+1), y=x/(x+1) and the y-axis is rotated about the y-axis through one complete revolution.

i) Use the method of cylindrical shells to show that the volume of the solid of revolution is given by:
aaaaaa1
V = 2pi∫{x(1-x)/(1+x)} dx
aaaaa0

ii) Write x(1-x)/(1+x) in the form ax + b + c/(x+1) to find the exact value of the above integral and hence the volume of the solid of revolution.

 

[Riviet]

= 2π{-x²/2 + 2x - 2ln(x+1)}₀¹

= 2π(1/2 + 2 - 2ln2)

Almost had it.

Don't forget a new question.

 

[Mountain.Dew]

Bah, I forgot both. Anyway:

Show that for n > 1

1ln(2/1) + 2ln(3/2) + ... + nln{(n+1)/n} = ln({n+1}ⁿ/n!)

1ln(2/1) + 2ln(3/2) + ... + nln{(n+1)/n}

LHS = ln 2 - ln 1 + 2ln3 - 2ln2 + 3ln4 - 3ln3 + 4ln5 - 4ln4 + ... + (n-1)ln(n) - (n-1)ln(n-1) + nln(n+1) - nln(n) = - ln1 - ln2 - ln3 - ln4 - .... - ln(n) + nln(n+1)

= nln(n+1) - (ln1 + ln2 + ln3 +...+ln(n)) = nln(n+1) - ln(1.2.3.4...n) = ln(n+1)ⁿ - ln(n!) = ln({n+1}ⁿ/n!) = RHS

 

[Mountain.Dew]

New question: tis an easy one

If a, b, c are roots of x^3 -9x + 9 = 0, show that

(a-1)(b-1)(c-1) = -1

4 marks.

 

[pLuvia]

x³-9x+9=0
(a-1)(b-1)(c-1)
=(ab-a-b+1)(c-1)
=abc-ac-bc+c-ab+a+b-1
=abc-(ac+ab+bc)+(a+b+c)-1
=-9-(-9)+0-1
=-1
As req'd

Have no quesiton at this point